Let assume $ X= \Omega \times (0,T) $ where $\Omega \subset \mathbb{R} $ is a bounded open set and $T>0$.

Now define the operator $ \mathcal{A} : C^{‎\sigma‎, \sigma‎/2‎}(‎X‎) \to C^{‎\sigma‎, \sigma‎/2‎‎}(‎X‎) $ in the following way:

$$ ‎‎‎\mathcal{A}‎‎(\phi)(t, ‎x)= \\ \nonumber‎ A_0(x) ‎‎\mathcal{B}(\phi)(t, ‎x, ‎0)+ \int_0^t ‎a(s, ‎x)‎b(s, ‎x) ‎| ‎\phi(s, ‎x)| ‎‎\mathcal{B}(\phi)(t,x,s) ‎ds $$

such that

$$ ‎\mathcal{B}(\phi)(t,x,s)=\chi‎_{[s, T]}(t) \, \exp\Big(-\int_s^t ‎\frac{a(\zeta,x) b(\zeta, x)}{c(\zeta, x)}\vert ‎\phi(\zeta, ‎x)\vert ‎\: ‎d‎\zeta‎\Big) ‎ \\ \nonumber‎ \times \exp \Big(-\int_s^t [d(\zeta, x)+‎e‎(‎\zeta‎, x)+f(\zeta, x) \chi‎_{‎\omega‎}]\: d‎\zeta ‎\Big)‎.$$


$$ a(t,x) , b(t,x) , c(t,x), d(t,x), e(t,x), f(t,x) \in C^{‎\sigma‎, \sigma‎/2‎} (‎\overline{\Omega} ‎\times ‎(0, ‎T)‎),$$

$\omega \subset \Omega$ and $A_0 \in C^{‎\sigma‎}(‎\overline{\Omega}‎).$

In solving a problem related to my thesis, for proving existence of a local solution by classical Contraction Mapping Theorem, I need to prove that the above operator is lipschitz with the standard norm of $C^{‎\sigma‎, \sigma‎/2‎‎}$, but I can't succeed.

Can someone help me to know whether this opeator can be lipschitz or not, in order to use other way to find a local solution instead of classical Contraction Mapping Theorem.

Thank you.

  • $\begingroup$ What values $\sigma$ can take? And shouldn't it be $C^{‎\sigma, ‎\sigma/2‎}(‎X‎)$ instead of $C^{‎\sigma‎/2, ‎\sigma‎}(‎X‎)$? $\endgroup$ – Andrew Jul 13 '18 at 18:47
  • $\begingroup$ @Andrew : Thanks Andrew, you are right. I have made mistake in typing. The correct Space is $C^{\sigma/2, \sigma}(X)$. The range of $\sigma$, in my problem belongs to $(0,1)$. $\endgroup$ – Hheepp Jul 14 '18 at 14:17
  • $\begingroup$ Apparently some assumptions on the coefficients are required: if $c(\zeta,x)$ is everywhere zero, then $\mathcal{B}$ is not even defined. Also, I fail to see why $b(\zeta, x)$ is needed, as it can be absorbed into $a(\zeta, x)$. $\endgroup$ – Mateusz Kwaśnicki Jul 16 '18 at 9:34
  • $\begingroup$ @ Mateusz Kwasnicki: Yes you are right. You can absorb b(.,.) into a(.,.). Also, you can put any assumtion on c(.,.), to make the integral meaningful. In other words, i want to see with any extra assumptions on coefficients, is it possible to show that the operator is contraction or not? $\endgroup$ – Hheepp Jul 17 '18 at 16:04

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